Solve left(x+3right)^2-left(2x-1right)^2=16

$${ \left(x+3 \right) }^{ 2 } - { \left(2x-1 \right) }^{ 2 } =16$$

Solve for x

$$x=2$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-\left(2x-1\right)^{2}=16$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$x^{2}+6x+9-\left(4x^{2}-4x+1\right)=16$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$x^{2}+6x+9-4x^{2}+4x-1=16$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+6x+9+4x-1=16$$

Combine 6x and 4x to get 10x.

$$-3x^{2}+10x+9-1=16$$

Subtract 1 from 9 to get 8.

$$-3x^{2}+10x+8=16$$

Subtract 16 from both sides.

$$-3x^{2}+10x+8-16=0$$

Subtract 16 from 8 to get -8.

$$-3x^{2}+10x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=10 ab=-3\left(-8\right)=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 10.

$$a=6 b=4$$

Rewrite -3x^{2}+10x-8 as \left(-3x^{2}+6x\right)+\left(4x-8\right).

$$\left(-3x^{2}+6x\right)+\left(4x-8\right)$$

Factor out 3x in the first and -4 in the second group.

$$3x\left(-x+2\right)-4\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(3x-4\right)$$

To find equation solutions, solve -x+2=0 and 3x-4=0.

$$x=2 x=\frac{4}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-\left(2x-1\right)^{2}=16$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$x^{2}+6x+9-\left(4x^{2}-4x+1\right)=16$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$x^{2}+6x+9-4x^{2}+4x-1=16$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+6x+9+4x-1=16$$

Combine 6x and 4x to get 10x.

$$-3x^{2}+10x+9-1=16$$

Subtract 1 from 9 to get 8.

$$-3x^{2}+10x+8=16$$

Subtract 16 from both sides.

$$-3x^{2}+10x+8-16=0$$

Subtract 16 from 8 to get -8.

$$-3x^{2}+10x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 10 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-3\right)\left(-8\right)}}{2\left(-3\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-3\right)\left(-8\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-10±\sqrt{100+12\left(-8\right)}}{2\left(-3\right)}$$

Multiply 12 times -8.

$$x=\frac{-10±\sqrt{100-96}}{2\left(-3\right)}$$

Add 100 to -96.

$$x=\frac{-10±\sqrt{4}}{2\left(-3\right)}$$

Take the square root of 4.

$$x=\frac{-10±2}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-10±2}{-6}$$

Now solve the equation x=\frac{-10±2}{-6} when ± is plus. Add -10 to 2.

$$x=-\frac{8}{-6}$$

Reduce the fraction \frac{-8}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{-10±2}{-6} when ± is minus. Subtract 2 from -10.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

The equation is now solved.

$$x=\frac{4}{3} x=2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-\left(2x-1\right)^{2}=16$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$x^{2}+6x+9-\left(4x^{2}-4x+1\right)=16$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$x^{2}+6x+9-4x^{2}+4x-1=16$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+6x+9+4x-1=16$$

Combine 6x and 4x to get 10x.

$$-3x^{2}+10x+9-1=16$$

Subtract 1 from 9 to get 8.

$$-3x^{2}+10x+8=16$$

Subtract 8 from both sides.

$$-3x^{2}+10x=16-8$$

Subtract 8 from 16 to get 8.

$$-3x^{2}+10x=8$$

Divide both sides by -3.

$$\frac{-3x^{2}+10x}{-3}=\frac{8}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{10}{-3}x=\frac{8}{-3}$$

Divide 10 by -3.

$$x^{2}-\frac{10}{3}x=\frac{8}{-3}$$

Divide 8 by -3.

$$x^{2}-\frac{10}{3}x=-\frac{8}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{8}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-\frac{8}{3}+\frac{25}{9}$$

Add -\frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{1}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{1}{3} x-\frac{5}{3}=-\frac{1}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$x=2 x=\frac{4}{3}$$